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php - SQL/PHP：在另一个表（多个表）上没有匹配项的联接

问题描述

我有四张桌子：

Personal_trainer(personaltrainerID, name, age, location)
Client(clientID, name, age, location)
Nutrition_Plan(NutritionplanID, personaltrainerID, clientID)
Training_Plan(TrainingplanID, personaltrainerID, clientID)

我试图表明，当私人教练创建营养/培训计划并将其分配给客户时，结果只会显示他们的特定客户，但他们在客户表中没有外键来识别教练（由于项目标准）

我想知道培训/营养计划的必要加入的 SQL 是什么。我已经尝试了很长一段时间，这是我的示例代码。

所需的输出是所有客户数据，前提是他们已被该特定培训师分配了培训或营养计划

在声明中，我遇到了 Bind 参数的问题，因此只有拥有客户端的用户才能看到他们的客户端。如果我使用指定的 ID，我可以获得退货！ 

    <?php 
            //code to search for a item from the database
            // user can enter any character to search for a value from the db
           if (isset($_POST['search']))
           {
               $valueToSearch = $_POST['ValueToSearch'];
               $query = "select * from client WHERE concat(`clientID`, `name`, `age`, `sex`, `weight`, `height`, `yearsExperience`, `goal`, `injuries`, 'email')like'%".$valueToSearch."%'";
               $search_result = filterTable($query);

           }
           else {
           $query = "select *
            from client
            where clientID in (select clientID from nutrition_plan where personaltrainerID=?)
            or clientID in (select clientID from training_plan where personalTrainerID=?)";
           $query->bind_param("i", $_POST[""]);
               $search_result = filterTable($query);
           }
           //code to filter the db
           function filterTable($query)
           {
               $connect = mysqli_connect("localhost:3308","root","","fypdatabase");
               $filter_Result = mysqli_query($connect, $query);
               return $filter_Result;
           }
           ?>

           <?php
           while($row = mysqli_fetch_array($search_result))
          { //display the details from the db in the table with option to delete or update entry 
            ?>
                    <tr>
                    <td><?php echo $row["clientID"]; ?></td>
                    <td><?php echo $row["name"]; ?></td>
                    <td><?php echo $row["age"]; ?></td>
                    <td><?php echo $row["sex"]; ?></td>
                    <td><?php echo $row["weight"]; ?></td>
                    <td><?php echo $row["height"]; ?></td>
                    <td><?php echo $row["yearsExperience"]; ?></td>
                    <td><?php echo $row["goal"]; ?></td>
                    <td><?php echo $row["injuries"]; ?></td>
                    <td><?php echo $row["email"]; ?></td>
                    <td> 
                        <a href="?Delete=<?php echo $row["clientID"]; ?>" onclick="return confirm('Are you sure?');">Delete</a>
                    </td>
                    <td>
                        <a href="updateClient.php?Edit=<?php echo $row["clientID"]; ?>" onclick="return confirm('Are you sure?');">Update</a>
                    </td>
                  </tr>
                  <?php

标签： phpmysqlsqljoinouter-join

解决方案

让我们看看我是否正确理解您的要求...

仅显示具有培训师 123 的营养或培训计划的客户：

select *
from client
where clientid in (select clientid from nutrition_plan where personaltrainerid = 123)
   or clientid in (select clientid from training_plan where personaltrainerid = 123);

向这些客户展示他们的所有计划（无论计划的培训师如何）：

select *
from client
join
(
  select 'nutrition' as kind, nutritionplanid as id, personaltrainerid, clientid
  from nutrition_plan
  union all
  select 'training' as kind, trainingplanid as id, personaltrainerid, clientid
  from training_plan
) plan using (clientid)
where clientid in (select clientid from nutrition_plan where personaltrainerid = 123)
   or clientid in (select clientid from training_plan where personaltrainerid = 123);

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